#### Provide solution for RD Sharma maths class 12 chapter 17  Maxima and Minima exercise 17.3  question 7

\begin{aligned} &a=3 \\ &b=-9 \& c \in R \end{aligned}

Hint:

Put,

\begin{aligned} &\mathrm{f}^{\prime}(-1)=0 \\ &\mathrm{f}^{\prime}(3)=0 \end{aligned}

Given:

$f(x)=x^{3}+a x^{2}+b x+C$

Explanation:

Differentiating f(x) with respect to x

\begin{aligned} f^{\prime}(x) &=3 x^{2}+2 a x+b+0 \\ &=3 x^{2}+2 a x+b \end{aligned}

\begin{aligned} &\text { Calculate } \mathrm{f}^{\prime}(\mathrm{x}) \text { at } \mathrm{x}=-1 \text { and } \mathrm{x}=3\\ &\begin{aligned} \mathrm{f}^{\prime}(-1) &=3(-1)^{2}+2 \mathrm{a}(-1)=\mathrm{b} \\ &=3-2 \mathrm{a}+\mathrm{b} \\ \mathrm{f}^{\prime}(3) &=3(3)^{2}+2 \mathrm{a}(3)+\mathrm{b} \\ &=27+6 \mathrm{a}+\mathrm{b} \end{aligned} \end{aligned}

Put and  $f^{\prime}(-1) =0$ and  $f^{\prime}(3)=0$ as they are the extremum values hence $f^{\prime}(x)=0$
\begin{aligned} &\therefore 3-2 \mathrm{a}+\mathrm{b}=0 \\ &7-2 \mathrm{a}+\mathrm{b}=-3-(1) \\ &27-6 \mathrm{a}+\mathrm{b}=20 \mid \\ &6 \mathrm{a}+\mathrm{b}=-27-(2) \\ &(1)-(2), \mathrm{wc} \text { get } \\ &-2 \mathrm{a}-6 \mathrm{a}=-3+27 \\ &-8 \mathrm{a}=2 \mathrm{y} \\ &\mathrm{a}=3 \\ &\text { Put } \mathrm{a}=3 \text { in }(1) \\ &-2(3)^{2}-\mathrm{b}=-3 \\ &\mathrm{~b}=-9 \end{aligned}